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Let I be the set of integers, N the set of non-negative integers; Np the set of non-positive integers ; E is the set of even integers and P is set of prime numbers. Then

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$I - N = N p$

$N ∆ N p = I - 0$

$N \cap N p = ϕ$

$E \cap P = ϕ$

answer is C.

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Detailed Solution

$N \cap N p = {0}$

$\begin{array}{l} I - N = {\dots, - 2, - 1} \\ N Δ N p = {N - N p) \cup (N p - N) \\ = {1, 2, 3 - -} \cup {\dots, - 3, - 2, - 1} \\ = I ~ {0} \end{array}$

and $E \cap P = {2} .$

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